Question

How many grams of pure naoh must be used to prepare 10.0 l of a solution that has a ph of 13? __________g?

Answer 1

40 g NaOH. You must use 40 g NaOH to prepare 10.0 L of a solution that has a pH of 13.

Step 1. Calculate the pOH of the solution

pOH = 14.00 – pH = 14.00 -13 = 1

Step 2. Calculate the concentration of NaOH

[NaOH] = [OH^(-)] = 10^(-pOH) mol/L = 10^(-1) mol/L = 0.1 mol/L

Step 3. Calculate the moles of NaOH

Moles of NaOH = 10.0 L solution × (0.1 mol NaOH/1 L solution) = 1 mol NaOH

Step 4. Calculate the mass of NaOH

Mass of NaOH = 1 mol NaOH × (40.00 g NaOH/1 mol NaOH) = 40 g NaOH

Answer 2

We must use 39.997 grams of NaOH to prepare 10.0 L of a solution that has a pH of 13.

To find the mass of NaOH, we need to calculate the number of moles.  

From the pH, we have:

$pH + pOH = 14$

$pOH = 14 - pH = 14 - 13 = 1$

The pOH is given by

$pOH = -log([OH^{-}])$

$[OH^{-}] = 10^{-pOH} = 10^{-1} = 0.10 M$

Now, we can find the number of moles with the volume of solution

$n = [OH^{-}]*V = 0.10 mol/L*10.0 L = 1.0 \:moles$

Finally, we can calculate the mass

$m = n*M = 1.0 \:moles*39.997 g/mol = 39.997 g$

Therefore, we need to use 39.997 g of NaOH.

Learn more here:

• https://brainly.com/question/8834103?referrer=searchResults

• https://brainly.com/question/1525823?referrer=searchResults

I hope it helps you!